\(\int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx\) [1005]

Optimal result

Integrand size = 31, antiderivative size = 57 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {A \sin (c+d x)}{a d}-\frac {(A-B) \sin ^2(c+d x)}{2 a d}-\frac {B \sin ^3(c+d x)}{3 a d} \]

[Out]

A*sin(d*x+c)/a/d-1/2*(A-B)*sin(d*x+c)^2/a/d-1/3*B*sin(d*x+c)^3/a/d

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2915, 45} \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {(A-B) \sin ^2(c+d x)}{2 a d}+\frac {A \sin (c+d x)}{a d}-\frac {B \sin ^3(c+d x)}{3 a d} \]

[In]

Int[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]

[Out]

(A*Sin[c + d*x])/(a*d) - ((A - B)*Sin[c + d*x]^2)/(2*a*d) - (B*Sin[c + d*x]^3)/(3*a*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2915

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a-x) \left (A+\frac {B x}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {\text {Subst}\left (\int \left (a A-(A-B) x-\frac {B x^2}{a}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d} \\ & = \frac {A \sin (c+d x)}{a d}-\frac {(A-B) \sin ^2(c+d x)}{2 a d}-\frac {B \sin ^3(c+d x)}{3 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\sin (c+d x) \left (6 A-3 (A-B) \sin (c+d x)-2 B \sin ^2(c+d x)\right )}{6 a d} \]

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x]),x]

[Out]

(Sin[c + d*x]*(6*A - 3*(A - B)*Sin[c + d*x] - 2*B*Sin[c + d*x]^2))/(6*a*d)

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {\frac {B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )}{d a}\)	\(45\)
default	\(-\frac {\frac {B \left (\sin ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (A -B \right ) \left (\sin ^{2}\left (d x +c \right )\right )}{2}-A \sin \left (d x +c \right )}{d a}\)	\(45\)
parallelrisch	\(\frac {\left (3 A -3 B \right ) \cos \left (2 d x +2 c \right )+B \sin \left (3 d x +3 c \right )+\left (12 A -3 B \right ) \sin \left (d x +c \right )-3 A +3 B}{12 d a}\)	\(58\)
risch	\(\frac {A \sin \left (d x +c \right )}{a d}-\frac {B \sin \left (d x +c \right )}{4 a d}+\frac {\sin \left (3 d x +3 c \right ) B}{12 a d}+\frac {\cos \left (2 d x +2 c \right ) A}{4 a d}-\frac {\cos \left (2 d x +2 c \right ) B}{4 a d}\)	\(85\)
norman	\(\frac {\frac {2 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 A \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 B \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {2 \left (6 A -B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {2 \left (6 A -B \right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {2 \left (3 A +2 B \right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}+\frac {2 \left (3 A +2 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\)	\(213\)

[In]

int(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

-1/d/a*(1/3*B*sin(d*x+c)^3+1/2*(A-B)*sin(d*x+c)^2-A*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {3 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (B \cos \left (d x + c\right )^{2} + 3 \, A - B\right )} \sin \left (d x + c\right )}{6 \, a d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/6*(3*(A - B)*cos(d*x + c)^2 + 2*(B*cos(d*x + c)^2 + 3*A - B)*sin(d*x + c))/(a*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 588 vs. \(2 (44) = 88\).

Time = 3.58 (sec) , antiderivative size = 588, normalized size of antiderivative = 10.32 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\begin {cases} \frac {6 A \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 A \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {12 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {6 A \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 B \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} - \frac {8 B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} + \frac {6 B \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a d \tan ^{6}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 9 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 3 a d} & \text {for}\: d \neq 0 \\\frac {x \left (A + B \sin {\left (c \right )}\right ) \cos ^{3}{\left (c \right )}}{a \sin {\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((6*A*tan(c/2 + d*x/2)**5/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*
x/2)**2 + 3*a*d) - 6*A*tan(c/2 + d*x/2)**4/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(
c/2 + d*x/2)**2 + 3*a*d) + 12*A*tan(c/2 + d*x/2)**3/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9
*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 6*A*tan(c/2 + d*x/2)**2/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 + d*x/2
)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*A*tan(c/2 + d*x/2)/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*tan(c/2 +
d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*B*tan(c/2 + d*x/2)**4/(3*a*d*tan(c/2 + d*x/2)**6 + 9*a*d*ta
n(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) - 8*B*tan(c/2 + d*x/2)**3/(3*a*d*tan(c/2 + d*x/2)**6 +
9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d) + 6*B*tan(c/2 + d*x/2)**2/(3*a*d*tan(c/2 + d*x/
2)**6 + 9*a*d*tan(c/2 + d*x/2)**4 + 9*a*d*tan(c/2 + d*x/2)**2 + 3*a*d), Ne(d, 0)), (x*(A + B*sin(c))*cos(c)**3
/(a*sin(c) + a), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {2 \, B \sin \left (d x + c\right )^{3} + 3 \, {\left (A - B\right )} \sin \left (d x + c\right )^{2} - 6 \, A \sin \left (d x + c\right )}{6 \, a d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/6*(2*B*sin(d*x + c)^3 + 3*(A - B)*sin(d*x + c)^2 - 6*A*sin(d*x + c))/(a*d)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.89 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=-\frac {2 \, B \sin \left (d x + c\right )^{3} + 3 \, A \sin \left (d x + c\right )^{2} - 3 \, B \sin \left (d x + c\right )^{2} - 6 \, A \sin \left (d x + c\right )}{6 \, a d} \]

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(2*B*sin(d*x + c)^3 + 3*A*sin(d*x + c)^2 - 3*B*sin(d*x + c)^2 - 6*A*sin(d*x + c))/(a*d)

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^3(c+d x) (A+B \sin (c+d x))}{a+a \sin (c+d x)} \, dx=\frac {\sin \left (c+d\,x\right )\,\left (6\,A-3\,A\,\sin \left (c+d\,x\right )+3\,B\,\sin \left (c+d\,x\right )-2\,B\,{\sin \left (c+d\,x\right )}^2\right )}{6\,a\,d} \]

[In]

int((cos(c + d*x)^3*(A + B*sin(c + d*x)))/(a + a*sin(c + d*x)),x)

[Out]

(sin(c + d*x)*(6*A - 3*A*sin(c + d*x) + 3*B*sin(c + d*x) - 2*B*sin(c + d*x)^2))/(6*a*d)